Let $h(x)=\dfrac{x^2+1}{2x^2-3x}$. $h'(x)=$
Answer: $h$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = h ′ ( x ) = d d x ( x 2 + 1 2 x 2 − 3 x ) = ( 2 x 2 − 3 x ) d d x ( x 2 + 1 ) − ( x 2 + 1 ) d d x ( 2 x 2 − 3 x ) ( 2 x 2 − 3 x ) 2 = ( 2 x 2 − 3 x ) ( 2 x ) − ( x 2 + 1 ) ( 4 x − 3 ) ( 2 x 2 − 3 x ) 2 = 4 x 3 − 6 x 2 − ( 4 x 3 − 3 x 2 + 4 x − 3 ) ( 2 x 2 − 3 x ) 2 = 4 x 3 − 6 x 2 − 4 x 3 + 3 x 2 − 4 x + 3 ( 2 x 2 − 3 x ) 2 = − 3 x 2 − 4 x + 3 ( 2 x 2 − 3 x ) 2 The quotient rule Differentiate ( 2 x 2 − 3 x ) & ( x 2 + 1 ) Expand \begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2+1}{2x^2-3x}\right) \\\\ &=\dfrac{(2x^2-3x)\dfrac{d}{dx}(x^2+1)-(x^2+1)\dfrac{d}{dx}(2x^2-3x)}{(2x^2-3x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(2x^2-3x)(2x)-(x^2+1)(4x-3)}{(2x^2-3x)^2}&&\gray{\text{Differentiate }(2x^2-3x)\text{ & }(x^2+1)} \\\\ &=\dfrac{4x^3-6x^2-(4x^3-3x^2+4x-3)}{(2x^2-3x)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{4x^3-6x^2-4x^3+3x^2-4x+3}{(2x^2-3x)^2} \\\\ &=\dfrac{-3x^2-4x+3}{(2x^2-3x)^2} \end{aligned} In conclusion, $h'(x)=\dfrac{-3x^2-4x+3}{(2x^2-3x)^2}$, or any other equivalent form.